Sorting columns and selecting top n rows in each group pandas dataframe

There are 2 solutions:

1.sort_values and aggregate head:

df1 = df.sort_values('score',ascending = False).groupby('pidx').head(2)
print (df1)

    mainid pidx pidy  score
8        2    x    w     12
4        1    a    e      8
2        1    c    a      7
10       2    y    x      6
1        1    a    c      5
7        2    z    y      5
6        2    y    z      3
3        1    c    b      2
5        2    x    y      1

2.set_index and aggregate nlargest:

df = df.set_index(['mainid','pidy']).groupby('pidx')['score'].nlargest(2).reset_index() 
print (df)
  pidx  mainid pidy  score
0    a       1    e      8
1    a       1    c      5
2    c       1    a      7
3    c       1    b      2
4    x       2    w     12
5    x       2    y      1
6    y       2    x      6
7    y       2    z      3
8    z       2    y      5

Timings:

np.random.seed(123)
N = 1000000

L1 = list('abcdefghijklmnopqrstu')
L2 = list('efghijklmnopqrstuvwxyz')
df = pd.DataFrame({'mainid':np.random.randint(1000, size=N),
                   'pidx': np.random.randint(10000, size=N),
                   'pidy': np.random.choice(L2, N),
                   'score':np.random.randint(1000, size=N)})
#print (df)

def epat(df):
    grouped = df.groupby('pidx')
    new_df = pd.DataFrame([], columns = df.columns)
    for key, values in grouped:
        new_df = pd.concat([new_df, grouped.get_group(key).sort_values('score', ascending=True)[:2]], 0)
    return (new_df)

print (epat(df))

In [133]: %timeit (df.sort_values('score',ascending = False).groupby('pidx').head(2))
1 loop, best of 3: 309 ms per loop

In [134]: %timeit (df.set_index(['mainid','pidy']).groupby('pidx')['score'].nlargest(2).reset_index())
1 loop, best of 3: 7.11 s per loop

In [147]: %timeit (epat(df))
1 loop, best of 3: 22 s per loop

a simple solution would be:

grouped = DF.groupby('pidx')

new_df = pd.DataFrame([], columns = DF.columns)

for key, values in grouped:

    new_df = pd.concat([new_df, grouped.get_group(key).sort_values('score', ascending=True)[:2]], 0)

hope it helps!